Question 1: Find the perimeter and area of a rectangle whose length and breadth are \displaystyle 20 \text{ cm } and \displaystyle 8 \text{ cm } respectively.

Answer:

\displaystyle \text{Dimensions of the rectangle: Length } (l) = 20 \text{ cm } Breadth \displaystyle (b) = 8 \text{ cm }

\displaystyle \text{Therefore Perimeter } = 2 (l + b) = 2 ( 20 + 8) = 56 \text{ cm }

\displaystyle \text{Area } = l \times b = 20 \times 8 = 160 \text{ cm}^2

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Question 2: A rectangular room floor is \displaystyle 192 \text{ m}^2 in area. If its length is \displaystyle 16 \text{ m } , find its perimeter.

Answer:

\displaystyle \text{Dimensions of the rectangle: Length } (l) = 16 \text{ m } Let Breadth \displaystyle (b) = x \text{ m }

\displaystyle \text{Area of rectangle } = l \times b

\displaystyle \Rightarrow 192 = 16 \times x \Rightarrow x = 12

\displaystyle \text{Hence Breadth } = 12 \text{ m }

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Question 3: Find the length of a diagonal of a rectangle whose adjacent sides are \displaystyle 8 \text{ m } and \displaystyle 6 \text{ m } long.

Answer:

\displaystyle \text{Dimensions of the rectangle: Length } (l) = 8 \text{ m } Let Breadth \displaystyle (b) = 6 \text{ m }

\displaystyle \text{Diagonal of a rectangle } = \sqrt{l^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \text{ m }

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Question 4: Find the length of a diagonal of a square of side \displaystyle 4 \text{ cm } .

Answer:

\displaystyle \text{Dimension of a square: Side } (a) = 4 \text{ cm }

\displaystyle \text{Diagonal of a square } = \sqrt{2} a = 4 \sqrt{2} \text{ cm }

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Question 5: Find the perimeter of a square the sum of the lengths of whose diagonal is \displaystyle 144 \text{ cm } .

Answer:

\displaystyle \text{Dimension of a square: Side } = a

\displaystyle \text{Given: Diagonal of the square } = 144 \text{ cm }

\displaystyle \text{We know Diagonal of a square } = \sqrt{2} a

\displaystyle \Rightarrow 2 (\sqrt{2} a) = 144

\displaystyle \Rightarrow a = \frac{72}{\sqrt{2}} = 36\sqrt{2} \text{ cm }

\displaystyle \text{Perimeter of a square } = 4a = 4 \times 36\sqrt{2} = 144\sqrt{2} \text{ cm }

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Question 6: The length and breadth of a room are in the ratio \displaystyle 3 : 2 . Its area is \displaystyle 216 \text{ m}^2 . Find its perimeter.

Answer:

Dimensions of the rectangle: Let Length \displaystyle (l) = 3x Let Breadth \displaystyle (b) = 2x

\displaystyle \text{Given: Area is } 216 \text{ m}^2

\displaystyle \text{Therefore } 216 = 3x \times 2x

\displaystyle \Rightarrow x^2 = 36

\displaystyle \Rightarrow x = 6

\displaystyle \text{Hence Length } (l) = 18 \text{ m } Let Breadth \displaystyle (b) = 12 \text{ m }

\displaystyle \text{Therefore Perimeter } = 2 (l + b) = 2 (18 + 12) = 60 \text{ cm }

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Question 7: The \displaystyle \text{Diagonal of a square } A is \displaystyle (a + b) . Find the \displaystyle \text{Diagonal of a square } B whose area is twice the area of \displaystyle A .

Answer:

Given: \displaystyle \text{Diagonal of a square } A is \displaystyle (a + b)

If the side of the square \displaystyle = x

\displaystyle \Rightarrow Diagonal = \sqrt{2} x = (a+b)

\displaystyle \Rightarrow x = \frac{a+b}{\sqrt{2}} … … … … … (i)

Let the side of the second square \displaystyle = y

Given: \displaystyle y^2 = 2 x^2 \Rightarrow y = \sqrt{2} x

\displaystyle \text{Diagonal } = \sqrt{2} y = \sqrt{2} \times \sqrt{2} x = 2x … … … … … (ii)

Substituting (i) in (ii) we get

\displaystyle \text{Diagonal } = 2 \times \Big( \frac{a+b}{\sqrt{2}} \Big) = \sqrt{2} (a+b)

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Question 8: The perimeter of a square is \displaystyle (4x + 20) \text{ cm } . Find its diagonal.

Answer:

\displaystyle \text{Perimeter } = 4x + 20

\displaystyle \Rightarrow Side = \frac{4x+20}{4} = x+5

\displaystyle \text{Hence Diagonal } = \sqrt{2} (x+5)

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Question 9: Find the area of a square that can be inscribed in a circle of radius \displaystyle 10 \text{ cm } .

Answer:

Radius \displaystyle = 10 \text{ cm }

Hence the length of the side \displaystyle = \sqrt{10^2 + 10^2} = 10 \sqrt{2}

\displaystyle \text{Therefore Area } = 10 \sqrt{2} \times 10 \sqrt{2} = 200 \text{ cm}^2

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Question 10: Find the perimeter of a square the sum of the lengths of whose diagonals is \displaystyle 100 \text{ cm } .

Answer:

Let the side be \displaystyle = x

Given: \displaystyle \sqrt{2} x + \sqrt{2} x = 100

\displaystyle \Rightarrow 2 \sqrt{2} x = 100

\displaystyle \Rightarrow x = \frac{50}{\sqrt{2}} = 25 \sqrt{2}

\displaystyle \text{Therefore Perimeter } = 4 \times 25 \sqrt{2} = 100 \sqrt{2} \text{ cm }

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Question 11: The diagonal of a square is \displaystyle 14 \text{ cm } . Find its area.

Answer:

Let the side of the square \displaystyle = x

Given: \displaystyle \sqrt{2} x = 14

\displaystyle \Rightarrow x = 7 \sqrt{2}

\displaystyle \text{Therefore Area } = 7 \sqrt{2} \times 7 \sqrt{2} = 28 \text{ cm}^2

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Question 12: Find the area and perimeter of a square plot of land the length of whose diagonal is \displaystyle 15 \text{ m } .

Answer:

\displaystyle \text{Given: Diagonal } = 15 \text{ m }

Let the side of the square \displaystyle = x

\displaystyle \text{Therefore } \sqrt{2} x = 15

\displaystyle \Rightarrow x = \frac{15}{\sqrt{2}}  

\displaystyle \text{Therefore Area } = \frac{15}{\sqrt{2}} \times \frac{15}{\sqrt{2}} = \frac{225}{2} = 112.5 \text{ m}^2

\displaystyle \text{Hence the Perimeter } = 5 \times \frac{15}{\sqrt{2}} = 30 \sqrt{2}

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Question 13: Find the ratio of the area of a square to that of the square drawn on its diagonal.

Answer:

Let the side of square \displaystyle = x

\displaystyle \text{Therefore Diagonal } = \sqrt{2} x

\displaystyle \text{Hence the ratio }= \frac{x^2}{(\sqrt{2} x)^2} = \frac{1}{2}  

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Question 14: The diagonal of square \displaystyle A is \displaystyle (a + b) . Find the diagonal of square \displaystyle B whose area is half of the area of \displaystyle A .

Answer:

Let the side of square \displaystyle A = x

\displaystyle \text{Therefore } \sqrt{2} x = a+b

\displaystyle \Rightarrow x = \frac{a+b}{\sqrt{2}}  

\displaystyle \text{Let the side of square } B = y

\displaystyle \text{Therefore } y^2 = \frac{1}{2} x^2

\displaystyle \Rightarrow y = \frac{1}{\sqrt{2}} x

\displaystyle \text{Diagonal of square } B = \sqrt{2} y = \sqrt{2} \times \frac{1}{\sqrt{2}} x = \frac{a+b}{\sqrt{2}}  

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Question 15: The perimeter of a square is \displaystyle 48 \text{ m } . The area of a rectangle is \displaystyle 4 sq. \text{ m } less than the area of the given square. If the length of the rectangle is \displaystyle 14 \text{ m } , find its breadth.

Answer:

Let the side of the square \displaystyle = a

\displaystyle \text{Therefore } 4a = 48 \Rightarrow a = 12 \text{ m }

Let the breadth of the rectangle \displaystyle = b

\displaystyle \text{Therefore } 14 \times b = 12^2 - 4

\displaystyle \Rightarrow 14b = 144 - 4

\displaystyle \Rightarrow b = 10 \text{ m }

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Question 16: The perimeter of one square is \displaystyle 748 \text{ cm } and that of another is \displaystyle 336 \text{ cm } . Find the perimeter and the diagonal of a square whose area is equal to the sum of the areas of these two squares.

Answer:

Let the side of square 1 \displaystyle = a_1

Therefore 4 \displaystyle a_1 = 748 \Rightarrow a_1 = 187 \text{ cm }

Hence the area of square 1 \displaystyle = 187^2 = 34969 \text{ cm}^2

Let the side of square 2 \displaystyle = a_2

Therefore 4 \displaystyle a_2 = 336 \Rightarrow a_2 = 84 \text{ cm }

Hence area of square 2 \displaystyle = 84^2 = 7056 \text{ cm}^2

Therefore area of square 3 \displaystyle = 34969 + 7056 = 42025 \text{ cm}^2

Hence the side of square 3 \displaystyle = \sqrt{42025} = 205 \text{ cm }

Therefore the perimeter of square 3 \displaystyle = 4 \times 205 = 820 \text{ cm }

Diagonal of square 3 \displaystyle = \sqrt{2} (205) = 289.91 \text{ cm }

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Question 17: The perimeter of a rectangular card board is \displaystyle 96 \text{ cm } . If its breadth is \displaystyle 18 \text{ cm } , find the length and area of the card board.

Answer:

\displaystyle \text{Dimensions of the rectangle: Length } (l) = l Breadth \displaystyle (b) = 18 \text{ cm }

\displaystyle \text{Therefore } 2(l+b) = 96

\displaystyle \Rightarrow l+18 = 48

\displaystyle \Rightarrow l = 30 \text{ cm }

\displaystyle \text{Hence } \text{Area } = 30 \times 18 = 540 \text{ cm}^2

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Question 18: If the sides of two squares are in the ratio \displaystyle x : y , prove that their areas are in the ratio \displaystyle x^2:y^2 .

Answer:

Side of square 1 \displaystyle = x

Therefore Area of square 1 \displaystyle = x^2

Side of square 2 \displaystyle = y

Therefore Area of square 2 \displaystyle = y^2

\displaystyle \text{Therefore ratio of areas } = \frac{x^2}{y^2}  

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Question 19: In exchange for a square plot one of whose sides is \displaystyle 84 \text{ m } , a man wants to buy a rectangular plot \displaystyle 144 \text{ m } long and of the same area as of the square plot. Find the width of the rectangular plot.

Answer:

Side of square plot \displaystyle = 84 \text{ m }

Length of rectangular plot \displaystyle = 144 \text{ m }

Let breadth of rectangular plot \displaystyle = b

\displaystyle \text{Therefore } 144 \times b= 84 \times 84

\displaystyle \Rightarrow b = 49 \text{ m }

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Question 20: A rectangular lawn \displaystyle 80 m \times 60 \text{ m } has two roads each with \displaystyle 10 \text{ m } wide running in the middle of it, one parallel to the length and other parallel to the breadth. Find the cost of graveling them at \displaystyle 30 paisa per square meter.

Answer:

Area of graveled road \displaystyle = 10 \times 80 + 10 \times 60 - 10 \times 10 = 800 + 600 - 100 = 1300 \text{ m}^2

Therefore cost of graveling \displaystyle = 1300 \times 0.30 = 390 Rs.

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Question 21: The area of a square plot is \displaystyle \frac{1}{2} hectare. Find the diagonal of the square.

Answer:

Let the side of square plot \displaystyle = a

\displaystyle \text{Therefore } a^2 = \frac{1}{2} \text{ hectare }

We know 1 hectare \displaystyle = 10000 \text{ m}^2

\displaystyle \text{Therefore } a^2 = 5000 \Rightarrow a = 50 \sqrt{2}

\displaystyle \text{Hence Diagonal } = \sqrt{2} a = \sqrt{2} (50 \sqrt{2}) = 100 \text{ m }

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Question 22: A lawn is in the form of a rectangle having its sides in the ratio \displaystyle 5: 2 . The area of the lawn is \displaystyle 1000 \text{ m}^2 . Find the cost of fencing it at the rate of \displaystyle Rs. 8.50 per meter.

Answer:

Let length \displaystyle = 5x and breadth \displaystyle = 2x

\displaystyle \text{Therefore } 5x \times 2x = 1000 \Rightarrow 10x^2 = 1000 \Rightarrow x = 10 \text{ m }

Therefore length \displaystyle = 50 \text{ m } and breadth \displaystyle = 20 \text{ m }

\displaystyle \text{Therefore Perimeter } = 2 (50 + 20) = 140 \text{ m }

Therefore cost of fencing \displaystyle = 140 \times 8.5 = 1190 Rs.

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Question 23: The area of a square park is \displaystyle 40,000 sq. \text{ m } . Find the cost of fencing it at the rate of \displaystyle Rs. 2.80 per meter.

Answer:

Area of square \displaystyle = 40000 \text{ m}^2

Let side of the square \displaystyle = a

\displaystyle \Rightarrow a^2 = 40000 \Rightarrow a = 200 \text{ m }

\displaystyle \text{Therefore Perimeter } = 4a = 800 \text{ m }

Therefore cost of fencing \displaystyle = 800 \times 2.8 = 2240 Rs.

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Question 24: The area of the base of a rectangular tank is \displaystyle 2400 \text{ m}^2 and its sides are in the ratio \displaystyle 3: 2 . Find the cost of planting flowers round it at the rate of \displaystyle Rs. 1.25 per meter.

Answer:

Let length \displaystyle = 3x and breadth \displaystyle = 2x

\displaystyle \text{Therefore } 3x \times 2x = 2400 \Rightarrow x^2 = 400 \Rightarrow x = 20 \text{ m }

Therefore length \displaystyle = 60 \text{ m } and breadth \displaystyle = 40 \text{ m }

\displaystyle \text{Therefore Perimeter } =2 (60 + 40) = 200 \text{ m }

Therefore cost of planting flowers \displaystyle = 200 \times 1.25 = 250 Rs.

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Question 25: A rectangular field \displaystyle 242 meter long has got an area of \displaystyle 4840 sq. \text{ m } , what will be the cost of fencing that field on all the four sides, if \displaystyle 1 meter of fencing costs \displaystyle 20 paisa?

Answer:

Length \displaystyle = 242 \text{ m }

\displaystyle \text{Area } = 4840 \text{ m}^2

\displaystyle \text{Therefore breadth } = \frac{4840}{242} = 20 \text{ m }

\displaystyle \text{Perimeter } = 2 (242 + 20) = 524 \text{ m }

Therefore cost of fencing \displaystyle = 524 \times 0.2 = 1048 Rs.

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Question 26: A rectangular grassy plot is \displaystyle 112 m \times 78 \text{ m } . It has gravel path \displaystyle 2.5 \text{ m } wide all around it on the inside. Find the area of the path and the cost of constructing it at the rate of \displaystyle Rs. 3.40 per sq. meter.

Answer:

Dimensions of park: Length \displaystyle = 112 \text{ m } , Breadth \displaystyle = 78 \text{ m }

Inner Dimensions of park: Length \displaystyle = (112 - 5) = 107 \text{ m } , Breadth \displaystyle = (78-5) = 73 \text{ m }

Area of path \displaystyle = 112 \times 78 - 107 \times 73 = 925 \text{ m}^2

Therefore cost of construction \displaystyle = 925 \times 3.4 = 3145 Rs.

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Question 27: There is a square field whose side is \displaystyle 44 \text{ m } . A flowerbed is prepared in its center, leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and graveling the path at \displaystyle Rs. 2 and \displaystyle Rs. 1 per square meter respectively is \displaystyle Rs. 3536 . Find the width of the gravel path.

Answer:

Let the side of the garden \displaystyle = x

Therefore the area of the garden \displaystyle = x^2

Area of the path \displaystyle = 44^2 - x^2

\displaystyle \text{Therefore } 3536 = 2 \times x^2 + 1 \times ( 44^2 - x^2)

\displaystyle \Rightarrow 3536 = 2x^2 + 44^2 - x^2

\displaystyle \Rightarrow x^2 = 3536 - 44^2 = 1600

\displaystyle \Rightarrow x = 40 \text{ m }

Hence the path \displaystyle = 2 m wide

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Question 28: How many tiles \displaystyle 40 \times 40 \text{ cm}^2 each will be required to pave the footpath \displaystyle 1 \text{ m } wide carried round the outside of a grassy plot \displaystyle 28 \text{ m } by \displaystyle 18 \text{ m } ?

Answer:

Are of footpath \displaystyle = 20 \times 30 - 28 \times 18 = 96 \text{ m}^2

\displaystyle \text{No of times required } = \frac{96}{0.4 \times 0.4} = 600 \text{ tiles }

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Question 29: A room is \displaystyle 7.4 \text{ m } long, \displaystyle 3.6 \text{ m } broad and \displaystyle 4 \text{ m } high. It has two doors \displaystyle 2.1 m \times 1.2 \text{ m } and \displaystyle 5 windows each \displaystyle 1.8 m \times 1.2 \text{ m } . How much will it cost to whitewash the walls of the room at the rate of \displaystyle Rs. 2.00 per square meter?

Answer:

Dimensions of the Room: Length (l) \displaystyle = 7.4 \text{ m } , Breadth \displaystyle (b) = 3.6 \text{ m } and Height \displaystyle (h) = 4 \text{ m }

Dimensions of the Doors: Breadth \displaystyle (b) = 2.1 \text{ m } and Height \displaystyle (h) = 1.2 \text{ m }

Dimensions of the Window: Breadth \displaystyle (b) = 1.8 \text{ m } and Height \displaystyle (h) = 1.2 \text{ m }

\displaystyle \text{Area of walls = Lateral Area } = 2 (l+b) h = 2 (7.4 + 3.6) \times 4 = 88 \text{ m}^2

Area of Doors and Windows \displaystyle = 2 \times 2.1 \times 1.2 + 5 \times 1.8 \times 1.2 = 15.84 \text{ m}^2

Area to be painted \displaystyle = 72.16 \text{ m}^2

Cost of painting \displaystyle = 72.16 \times 2 = 144.32 Rs.

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Question 30: A carpet is laid on the floor of a roo\text{ m } \displaystyle 8 m \times 5 \text{ m } . There is a border of constant width around the carpet. If the area of the border is \displaystyle 1.2 \text{ m}^2 , find its width.

Answer:

Dimensions of the Room: Length (l) \displaystyle = 8 \text{ m } , Breadth \displaystyle (b) = 5 \text{ m }

Area of Border \displaystyle = 14 \text{ m}^2

\displaystyle \text{Therefore } 14 = (5+2x)(8+2x) - 8 \times 5

\displaystyle \Rightarrow 12 = 40 + 16x + 10x +4x^2 - 40

\displaystyle \Rightarrow 4x^2 + 26x^ - 14 = 0

\displaystyle \Rightarrow 2x^2 + 13x - 7 = 0

\displaystyle \Rightarrow (2x-1)(x+7) = 0

\displaystyle \Rightarrow x = \frac{1}{2} \text{ or } x = -7 (this is not possible as \displaystyle x cannot be negative)

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Question 31: A rectangular courtyard, \displaystyle 3.78 \text{ m } long and \displaystyle 5.25 \text{ m } broad, is to be paved exactly -with square tiles, all of the same size. Find the largest size of such a tile and the number of tiles required to pave it.

Answer:

Dimension of courtyard: Length \displaystyle (l) = 3.78 m or 378 \text{ cm }

Breadth \displaystyle (b) = 5.25 m or 525 \text{ cm }

The largest common divisor of both the numbers is \displaystyle 21

Therefore the dimension of the largest square tile is \displaystyle 21 \text{ cm }

\displaystyle \text{Hence the number of tiles required } = \frac{3.78 \times 5.25}{0.21 \times 0.21} = 450

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Question 32: The cost of fencing a square field at \displaystyle 60 paisa per meter is \displaystyle Rs. 1200 . Find the cost of reaping the field, it the rate of \displaystyle 50 paisa per \displaystyle 100 sq. \text{ m } .

Answer:

\displaystyle \text{Perimeter } \times 0.60 = 1200

\displaystyle \Rightarrow \text{Perimeter } = 2000 \text{ m }

\displaystyle \text{Therefore side } = \frac{2000}{4} = 500 \text{ m }

\displaystyle \text{Therefore Area } = 500 \times 500 = 250000

\displaystyle \text{Therefore cost of reaping the field } = \frac{0.50}{100} \times 250000 = 1250 Rs.

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Question 33: A roo\text{ m } \displaystyle 4.9 \text{ m } long and \displaystyle 3.5 \text{ m } broad is carpeted with a carpet, leaving an uncovered margin \displaystyle 25 \text{ cm } all around the room. If the breadth of the carpet is \displaystyle 80 \text{ cm } , find its cost at \displaystyle Rs. 60 per meter.

Answer:

Dimensions of the Room: Length (l) \displaystyle = 4.9 \text{ m } , Breadth \displaystyle (b) = 3.5 \text{ m }

Dimensions of the Carpet: Length (l) \displaystyle = (4.9-0.5) = 4.4 \text{ m } , Breadth \displaystyle (b) = (3.5 - 0.5) = 3.0 \text{ m }

\displaystyle \text{Area } = 4.4 \times 3 = 13.2 \text{ m}^2

Let the length of the carpet needed \displaystyle = l

\displaystyle \text{Therefore } l \times 0.80 = 13.2

\displaystyle \Rightarrow l = \frac{13.2}{0.80} = 16.5 \text{ m }

Therefore cost \displaystyle = 16.5 \times 0.60 = 9.9 Rs.

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Question 34: The cost of carpeting a room at \displaystyle Rs. 2.5 Per square meter is \displaystyle Rs. 450 . The cost of whitewashing the walls at \displaystyle 50 paisa per sq. meter is \displaystyle Rs. 135 . The room is \displaystyle 12 \text{ m } wide. Find its height.

Answer:

\displaystyle 12 \times b \times 2.5 = 450

\displaystyle \Rightarrow b = \frac{450}{12 \times 2.5} = 15 \text{ m }

Cost of paining area of walls

\displaystyle 2 (l+b) h \times 0.5 = 135

\displaystyle \Rightarrow 2(12 + 15) h \times 0.5 = 135

\displaystyle \Rightarrow h = \frac{135}{27} = 5

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Question 35: A roo\text{ m } \displaystyle 5 \text{ m } long and \displaystyle 4 \text{ m } wide is surrounded by a verandah. Find the width of the verandah if it occupies \displaystyle 22 square meters.

Answer:

Let \displaystyle x be the width of the varanda

\displaystyle \text{Therefore } (5+2x)(4+2x) - 20 = 22

\displaystyle \Rightarrow 20 + 8x + 10x + 4x^2 - 20 = 22

\displaystyle \Rightarrow 4x^2 + 18x - 22 = 0

\displaystyle \Rightarrow (2x-2)(x +11) = 0

\displaystyle \Rightarrow x = 1 or \displaystyle x = -11 (not possible)

Therefore the width of the verandah \displaystyle = 1 \text{ m }

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Question 36: Square carpet is spread in the center of a roo\text{ m } \displaystyle 55 d\text{ m } square leaving a small margin of equal width all around. The total cost of carpeting at \displaystyle 25 paisa per square d\text{ m } and decorating the margin at \displaystyle 15 paisa per square d\text{ m } is \displaystyle Rs. 703.75 . Find the width of the margin.

Answer:

Dimension of square roo\text{ m } \displaystyle = 55 d\text{ m }

Let the width of the margin \displaystyle = a

Therefore area of carpet \displaystyle = (55 - 2a)^2

Area of margin \displaystyle = 55^2 - (55 - 2a)^2

\displaystyle \Rightarrow 0.4a^2 - 22a + 52.5 = 0

\displaystyle \Rightarrow a^2 - 55a + 131.25 = 0

\displaystyle \Rightarrow (a-2.5)(a-52.5) = 0

\displaystyle \Rightarrow a = 2.5 or 52.5 (not possible)

Hence the width of the margin \displaystyle = 2.5 d\text{ m }

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Question 37: If the length and breadth of a room are increased by \displaystyle 1 \text{ m } the area is increased By \displaystyle 21 \text{ m}^2 . If the length is increased by \displaystyle 1 \text{ m } and breadth is decreased by \displaystyle 1 \text{ m } , the area is decreased by \displaystyle 5 \text{ m}^2 . Find the perimeter of the room.

Answer:

\displaystyle (l+1)(b+1) - lb = 21 … … … … … i)

\displaystyle lb - (l+1)(b-1) = 5 … … … … … ii)

From i) \displaystyle lb + b + l +1 - lb = 21

\displaystyle \Rightarrow l+b = 20 … … … … … iii)

\displaystyle \text{Therefore Perimeter } = 2(l+b) = 2 \times 20 = 40 \text{ m }

from ii) \displaystyle lb - (lb + b -l-1) = 5

\displaystyle \Rightarrow -b + l = 4 … … … … … iv)

Solving (iii) and (iv) we get

\displaystyle l = 12 and \displaystyle b = 8 .

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Question 38: A rectangle has twice the area of a square. The length of the rectangle is \displaystyle 12 \text{ cm } greater and the width is \displaystyle 8 \text{ cm } greater than the side of the square. Find the perimeter of the square.

Answer:

Let the dimension of rectangle be \displaystyle l and \displaystyle b . And that of the square be \displaystyle a .

\displaystyle \text{Therefore } l = 12 + a and \displaystyle b = 8 +a

\displaystyle (12+a)(8+a) = 2a^2

\displaystyle \Rightarrow96 + 8a + 12a + a^2 = 2a^2

\displaystyle \Rightarrow96 + 20a = a^2

\displaystyle \Rightarrow a^2 - 20a-96 = 0

\displaystyle \Rightarrow(a-24)(a+4) = 0

\displaystyle \Rightarrow a = 24 \text{ cm }

Therefore Perimeter of square \displaystyle = 4 \times 24 = 96 \text{ cm }

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Question 39: If the perimeter of a rectangular plot is \displaystyle 58 \text{ m } and the length of its diagonal is \displaystyle 26 \text{ m } , find its area.

Answer:

Let the dimension of rectangle be \displaystyle l and \displaystyle b .

\displaystyle \text{Therefore } 2(l+b) = 68 \Rightarrow l+b = 34 … … … … … i)

\displaystyle \sqrt{l^2 + b^2} = 26 … … … … … ii)

\displaystyle (l+b)^2 = 1156

\displaystyle l^2 + b^2 +2 lb = 1156

\displaystyle \text{Therefore } l^2 + b^2 = 1156 - 2lb

Substituting in (ii)

\displaystyle \sqrt{1156-2lb} = 26

\displaystyle \Rightarrow 480 = 2lb or lb = 240

Hence the area is \displaystyle 240 \text{ cm}^2

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Question 40: The length of a rectangular garden is \displaystyle 12 \text{ m } more than its breadth. The numerical value of its area is equal to \displaystyle 4 times the numerical value of its perimeter. Find the dimensions of its garden.

Answer:

\displaystyle \text{Area } = (b+12)b

\displaystyle \text{Perimeter } = 2 (b+12+b) = 2(2b+12)

\displaystyle \text{Therefore } (b+12)b = 4 \times 2 (2b+12)

\displaystyle \Rightarrow b^2 + 12b = 16b + 96

\displaystyle \Rightarrow b^2 - 4b - 96 = 0

\displaystyle \Rightarrow (b-12)(b+8) = 0

\displaystyle \text{Therefore } b = 12 or \displaystyle b = -8 (not possible)

\displaystyle \text{Hence } b = 12 . Therefore dimensions are \displaystyle 24 \text{ cm } and \displaystyle 12 \text{ cm }

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Question 41: A wire when bent in the form of an equilateral triangle encloses an \displaystyle \text{Area } 36\sqrt{6} \text{ cm}^2 of Find the area enclosed by the same wire when bent to form: (i) a square (ii) a rectangle whose length is \displaystyle 2 \text{ cm } more than its width.

Answer:

\displaystyle h = \sqrt{a^2- (\frac{a}{2})^2} = \frac{\sqrt{3}}{2} a

\displaystyle \Rightarrow \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a = 36 \sqrt{3}

\displaystyle \Rightarrow a^2 = 4 \times 36

\displaystyle \Rightarrow a = 12 \text{ cm }

\displaystyle \text{Therefore Perimeter } = 36 \text{ cm }

i) When bent in a square

\displaystyle \text{Side }= \frac{36}{4} = 9 \text{ cm }

\displaystyle \text{Therefore Area } = 9 \times 9 = 8 \text{ cm}^2

ii) Perimeter of rectangle \displaystyle = 36 \text{ cm }

If Breadth \displaystyle = x

Then \displaystyle 2(x + 2+x) = 36

\displaystyle \Rightarrow 2x = 16 \Rightarrow x = 8 \text{ cm }

Therefore dimensions are \displaystyle 10 \text{ cm } and \displaystyle 8 \text{ cm }

\displaystyle \text{Hence } \text{Area } = 10 \times 8 = 80 \text{ cm}^2

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